In (a.o.) a Newtonian telescope, there is an area on the primary that lies in the shadow of the secondary. Since there is a shadow on the primary, there also might be a shadow of the secondary on itself.
To ease collimation, people like to mark the center of the primary and or secondary. It is thus an often heard question "Can I do so without affecting the optical performance ?". The answer is : It depends ! (surprise :-)
In order to be able to answer this questing, one needs to specify the area in the focal plane centered on the focal axis, where a marking must have no impact. In the formula's below, this area is called "unaffected area" and symbolized with the letter "k"
For those of you that allergic to formula's there is good news and bad news, the good news is that if the unaffected area is chosen to be the 100 % illuminated area, then you can mark the primary mirror without hesitation. The bad news is that the secondary may have a shadow, but to be sure you have to use the formula's. The examples below plus additional examples not shown here all indicated that there was no shadow on the secondary at all (under the above mentioned condition). The only way to have a shadow on the secondary was to assume a very small value for "k" and a very fast primary (f/4). So we are left with the conclusion that the secondary on the common newtonians should not be marked.
The shape of the shadow on the primary is not a circle because the secondary is placed at a 45 degree angle (w.r.t. the optical axis). This also offsets the center of the shadow in a direction opposite to the focuser. For the secondary it is also offset away from the focuser.
The formula's below are valid for the direction parallel to the short axis of the secondary. They can also be used for the direction parallel to the long axis of the secondary, but then you'll have to adjust the value of "h" (distance between primary and secondary on the optical axis). For the shadow generated by the secondary part at the side of the focuser use the value h + n for "h" and for the shadow generated by the secondary part opposite the focuser use h - n for "h". ("n" is half the short axis of the secondary) If you have an offset applied to the center of the secondary, then the values become h' = h + (n - HD) and h' = h - (n + HD) with HD being the horizontal (= vertical) displacement of the secondary. Also the value of "n" itself must be corrected in the same way. (i.e. n'= n - HD, and n' = n + HD)
For a view of figure used to derive the formula's, click here.
Let's assume the radius of the circle of this area is "k", then the diameter of the area on the primary that is not used due to the secondary shadow is approximately:
It seems reasonable to make the area that should not be affected by a dot on the primary equal to the fully illuminated area. Go there for the relation between "n" and "k".
The above formula is an approximation, but a very good one. The error makes the value for "c" actually a little bit smaller than it should be. Just in case you are interested in the "real" value, fill in the value "c" in the formula's below:
Since the shadow on the primary is reflected back onto the secondary, it stands to reason to wonder how big this area is. The approximation is given here:
If the value "cs" gets negative, then there is no area on the secondary that has a shadow. I.e. the unaffected area uses the whole surface of the secondary.
And again for those who like the formula for the "real" value:
In the examples below, "k" is equal to the fully illuminated area, all values are in [mm] (= 1/25 Inch) and there is no secondary offset applied. Values are rounded.
k = 5 mm
h = f - Radius_of_primary - 50 mm
Diameter: 150 mm (6") 200 mm (8") 250 mm (10")
C Cs C Cs C Cs
f/#
4 15.7 -0.7 18.8 -0.5 22.0 -0.3
6 10.4 -2.9 12.5 -2.8 14.6 -2.7
8 7.8 -3.7 9.4 -3.7 10.9 -3.6
10 6.3 -4.1 7.5 -4.1 8.8 -4.0
The secondary was supposed to have the correct value to achieve a fully illuminated area of exactly 5 mm.